Selection Rule delta_I = 1/2 in the strangeness-changing weak currents

  • Thread starter Thread starter zq460
  • Start date Start date
  • Tags Tags
    Currents Weak
zq460
Messages
4
Reaction score
0
I read in a paper that delta_I = 1/2 rule of the strangeness-changing weak current implies the ratio of cross section of proton->Sigma0 process to neutron->Sigma- process to be 1/2. I do not understand what is this selection rule and how does it effect the cross section ratio of the 2 processes. Is delta_I = 1/2 rule is the change in the isospin of initial and final particle which has to be 1/2 for these processes? Will be great if somebody could explain. Thank you.
 
Physics news on Phys.org
zq460, Could you be more specific about what reactions you have in mind? The only references I can find about a ΔI = 1/2 rule apply to K → ππ decay.
 
yes, I should have posted the reference paper. Sorry my first time ever on this forum. Here you can see an old paper http://nngroup.physics.sunysb.edu/~nngroup/misc/Documents/NeutrinoReactionsAtAcceleratorEnergies.pdf. The delta_I = 1/2 rule is mentioned just before equation (3.40) on page 317.

I am talking about the cabbibo suppressed 'charged current processes', where when a neutrino interacts with a nucleon, converts to a lepton producing a W-, which converts a u-quark in the nucleon to a s-quark. So this is a strangness changing process (delta_S = 1) where a 'proton converts to a Sigma0' or a 'neutron converts to a Sigma-', both processes are written in equation (3.39) and their cross section ratio is written to be 1/2 accodording to delta_I = 1/2 rule in equation (3.40). I do not understand the rule and also how it effects the ratio of the cross sections of the 2 processes?

Thanks much for the help!
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top